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# A certain musical scale has has 13 notes, each having a

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A certain musical scale has has 13 notes, each having a [#permalink]  01 Jan 2004, 22:34
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A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Feb 2013, 05:01, edited 1 time in total.
Edited the question and added OA.
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Re: OG PS #434-- musical scale [#permalink]  02 Jan 2004, 01:49
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stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)

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Let k be the constant.

Hence, f2/f1=k. given, f1=440, f13=highest frequency=2*440=880.

Also, f13=440*k^12

therefore, 440*k^12=880 , or k^12=2, or (k^6)^2=2 or k^6=root 2.

f7=440k^6= 440*root2.
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A quick gut check on this problem is that every choice but A is way more than 880. Logically the 7th note must have a smaller value(the problem also states 'lower' frequency) than than the 13th. If crunched for time, good guess would be A.
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Re: Took me a white to get it [#permalink]  23 Nov 2004, 17:54
toddmartin wrote:
It took me about 3.5 minutes to get the answer, even though I did get it right. I had to think about it for a few seconds then write down what I knew in order to see the solution. How difficult is this problem in relation to the ones I'm likely to see on the test if I'm aiming for 710-750?

Do not worry too much. I think this type of question does not appear unless you have 51 in quant. You can definitely break the 700 barrier without reaching 51 quant as long as you have a good verbal score also.
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I had no idea how to start it so I just
wrote down what I had and tried reorganizing
it.

440 * k = 2nd tone
440 * k * k = 3rd tone
.
.
440 * k ^6 = 7th tone.

At this point, I was wondering why none of
power, but after I wrote 440 (k^12) = 880
and started simplifying, I stumbled on the answer.

Congratulations to ruhi160184.
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Re: OG PS #434-- musical scale [#permalink]  09 Nov 2011, 03:00
Praetorian wrote:
stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)

The question stem says that : the ratio of frequency to next higer frequency is fixed constant ...

Doesnt that mean F1/F2 = k

F2/F3 = k^2 and something like that .....???and not F2/F1 = k which is F2 = kF1???
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Re: [#permalink]  09 Nov 2011, 03:03
ruhi wrote:
Let k be the constant.

Hence, f2/f1=k. given, f1=440, f13=highest frequency=2*440=880.

Also, f13=440*k^12

therefore, 440*k^12=880 , or k^12=2, or (k^6)^2=2 or k^6=root 2.

f7=440k^6= 440*root2.

The question stem says that : the ratio of frequency to next higer frequency is fixed constant ...

Doesnt that mean F1/F2 = k

F2/F3 = k^2 and something like that ..... ??? and not F2/F1 = k which is F2 = kF1???
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Re: A certain musical scale has 13 notes, each having a [#permalink]  09 Nov 2011, 19:01
Slightly different way of approaching it...

Given

$$n_1 = 440$$

$$n_{13} = 880$$

$$n_i = 440(1+k)^{i-1}$$

Solve for $$n_7$$

Given that: $$n_7 = 440(1+k)^{6}$$

$$n_{13} = 440(1+k)^{12}$$

$$880 = (440(1+k)^{6})(1+k)^{6}$$

$$440*880 = (440(1+k)^{6})(440(1+k)^{6})$$

$$2*440^2 = (n_7)^2$$

$$n_7 = \sqrt{2}(440)$$
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Re: On another thread, someone with a very good GMAT score said [#permalink]  09 Nov 2011, 21:27
Siddhans,

It says the ratio of a frequency to the next higher frequency is a constant. f2/f1=f3/f2=f4/f3=.....=fixed number. This is an example of a geometric progression.

Here is a video explanation:

http://www.gmatquantum.com/og10-journal ... ition.html

Dabral
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Re: Re: [#permalink]  10 Nov 2011, 21:32
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siddhans wrote:
ruhi wrote:
Let k be the constant.

Hence, f2/f1=k. given, f1=440, f13=highest frequency=2*440=880.

Also, f13=440*k^12

therefore, 440*k^12=880 , or k^12=2, or (k^6)^2=2 or k^6=root 2.

f7=440k^6= 440*root2.

The question stem says that : the ratio of frequency to next higer frequency is fixed constant ...

Doesnt that mean F1/F2 = k

F2/F3 = k^2 and something like that ..... ??? and not F2/F1 = k which is F2 = kF1???

Responding to a pm:

The statement, "the ratio of a frequency to the next higher frequency is a fixed constant." means that the ratio of two consecutive frequencies is always the same.
It doesn't matter how you write it.
You can say F1/F2 = F2/F3 = F3/F4 = ... = F12/F13 = k
You can also say F2/F1 = F3/F2 = F4/F3 = ... = F13/F12 = k
The two constants are different. Your k will be reciprocal of each other in the two cases. You can follow any approach. You will get the value of k accordingly.

Mind you, F2/F3 is also k, not k^2. The ratio remains constant. It is a geometric progression. The ratio between any two consecutive values is always the same.
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Re: OG PS #434-- musical scale [#permalink]  22 Feb 2013, 03:38
Praetorian wrote:
stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)

Why do you multiply and not add?

Like
1. 440
2. 440 + k and so on?
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Re: A certain musical scale has has 13 notes, each having a [#permalink]  22 Feb 2013, 04:48
Lowest frequency = 440
Highest frequency = 880
Lowest frequency (n) ^12 = Highest frequency
N^12 = 2 ---------------- 1
7th note = Lowest Frequency x (n)^6
7th note = 440 x (2)^6/12
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Re: OG PS #434-- musical scale [#permalink]  22 Feb 2013, 05:12
Expert's post
karmapatell wrote:
Praetorian wrote:
stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)

Why do you multiply and not add?

Like
1. 440
2. 440 + k and so on?

Because we are given that the ratio of a frequency to the next higher frequency is a fixed constant: F2/F1=k.
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Re: OG PS #434-- musical scale [#permalink]  22 Feb 2013, 09:29
karmapatell wrote:
Praetorian wrote:
stoolfi wrote:

On another thread, someone with a very good GMAT score said he couldn't really understand this question. It is posted for your perusal:

A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

let the constant be k

F1 = 440

F2 = 440k

F3 = 440 k * k = 440 * k^2

F13= 440 * k^12

we know F13 = 2 *F1 = 2 * 440 = 880

880/440 = k^12

k = twelfth root of 2

for F7...

F7 = 440 * k^6 ( as we wrote for F2 and F3)
F7 = 440 * (twelfth root of 2) ^ 6

F7 = 440 * sqrt (2)

Why do you multiply and not add?

Like
1. 440
2. 440 + k and so on?

"For each of the 12 lower frequencies, the ratio of a frequency to
the next higher frequency is a fixed constant"

The question says that the ratio of two consecutive frequencies is constant. Hence we multiply.
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Re: A certain musical scale has has 13 notes, each having a [#permalink]  22 Feb 2013, 12:59
Shoudln't "the ratio of a frequency to the next higher frequency is a fixed constant" be interpreted as f1/f2 = k instead of f2/f1=k?
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Re: A certain musical scale has has 13 notes, each having a [#permalink]  22 Feb 2013, 21:51
Let's take a simple example

Series : 2 4 6 8
Now going forward the multiplying factor is 2 and backward the factor is 1/2

Hence if you go backwards in the question the factor will be 1/2 and still you will get the same answer.

Posted from my mobile device
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Re: A certain musical scale has has 13 notes, each having a [#permalink]  23 Jun 2013, 14:04
Since the ratio is constant it is a geometric progression

7th term is the middle term of the series

mean of the geometric series given by sqrt(a*b)

a= first term of the series = 440
b= last term of the series = 2*440

Mean = 7th term = sqrt(2*440*440) = 440*sqrt(2)

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Re: A certain musical scale has has 13 notes, each having a [#permalink]  12 Jul 2013, 21:40
Pushpinder wrote:
Lowest frequency = 440
Highest frequency = 880
Lowest frequency (n) ^12 = Highest frequency
N^12 = 2 ---------------- 1
7th note = Lowest Frequency x (n)^6
7th note = 440 x (2)^6/12

Pushpinder Ji I couldn't understand from here. Can u tell me please

Last edited by Bunuel on 12 Jul 2013, 23:28, edited 2 times in total.
Edited.
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Re: A certain musical scale has has 13 notes, each having a [#permalink]  12 Jul 2013, 23:38
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dasikasuneel wrote:
Pushpinder wrote:
A certain musical scale has has 13 notes, each having a different frequency, measured in cycles per second. In the scale, the notes are ordered by increasing frequency, and the highest frequency is twice the lowest. For each of the 12 lower frequencies, the ratio of a frequency to the next higher frequency is a fixed constant. If the lowest frequency is 440 cycles per second, then the frequency of the 7th note in the scale is how many cycles per second?

A. 440 * sqrt 2
B. 440 * sqrt (2^7)
C. 440 * sqrt (2^12)
D. 440 * the twelfth root of (2^7)
E. 440 * the seventh root of (2^12)

Lowest frequency = 440
Highest frequency = 880
Lowest frequency (n) ^12 = Highest frequency
N^12 = 2 ---------------- 1
7th note = Lowest Frequency x (n)^6
7th note = 440 x (2)^6/12

Pushpinder Ji I couldn't understand from here. Can u tell me please

1st = $$440$$
2nd = $$440k$$
3rd = $$440k^2$$
...
7th = $$440k^6$$
...
13th = $$440k^{12}=2*440=880$$ --> $$440k^{12}=880$$ --> $$k^{12}=2$$ --> $$k=\sqrt[12]{2}$$.

Thus, 7th = $$440k^6=440(\sqrt[12]{2})^6=440\sqrt{2}$$.

Hope it's clear.
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Re: A certain musical scale has has 13 notes, each having a   [#permalink] 12 Jul 2013, 23:38

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